# Is the geometric realization of a pointed category contractible?

Given a pointed category $X$, is $|N(X)|$ contractible ($N(X)$ is the nerve of $X$)? Or are there counter-examples?

• What is your definition of pointed category? A category with a zero object? – Eric Wofsey Sep 17 '18 at 20:39
• @EricWofsey Yes – 6666 Sep 17 '18 at 20:48

Any category that has either a terminal object or an initial object has contractible nerve. This follows from the fact that the nerve functor turns natural transformations into homotopies, and thus turns adjoint functors into inverse homotopy equivalences (since the unit and counit of the adjunction become homotopies to the identity). A terminal object is exactly a right adjoint to the unique functor $X\to 1$, and so gives a homotopy equivalence $|N(X)|\simeq |N(1)|$ and hence shows $|N(X)|$ is contractible since $|N(1)|$ is a one-point space. Similarly, an initial object is a left adjoint to $X\to 1$.
• "nerve functor turns natural transformations into homotopies" do you mean $|\cdot|: \textbf{Cat}\to\textbf{Top}$ turns natural transformations into homotopies? where $X\mapsto |N(X)|$. – 6666 Sep 17 '18 at 21:10
• Yes. Or already $N:\mathbf{Cat}\to \mathbf{sSet}$ turns natural transformation into homotopies of simplicial sets, and then of course geometric realization turns homotopies of simplicial sets into topological homotopies. – Eric Wofsey Sep 17 '18 at 21:22
• I don't know a reference off the top of my head. The easiest way to prove it is to observe that a natural transformation between two functors $C\to D$ is the same as a functor $C\times I\to D$ whose restrictions to $C\times\{0\}$ and $C\times\{1\}$ are your two functors, where $I$ is the category with two objects $0$ and $1$ and one non-identity morphism $0\to 1$. Then, the nerve functor sends $I$ to $\Delta^1$ and preserves products, so you get a homotopy $N(C)\times \Delta^1\to N(D)$. – Eric Wofsey Sep 17 '18 at 21:55